Problem: You have found the following ages (in years) of 5 turtles. Those turtles were randomly selected from the 48 turtles at your local zoo: $ 30,\enspace 110,\enspace 21,\enspace 87,\enspace 101$ Based on your sample, what is the average age of the turtles? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 48 turtles, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $5$ samples and divide by $5$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\overline{x}} = \dfrac{30 + 110 + 21 + 87 + 101}{{5}} = {69.8\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {1584.04} + {1616.04} + {2381.44} + {295.84} + {973.44}} {{5 - 1}} $ {s^2} = \dfrac{{6850.8}}{{4}} = {1712.7\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{1712.7\text{ years}^2}} = {41.4\text{ years}} $ We can estimate that the average turtle at the zoo is 69.8 years old. There is also a standard deviation of 41.4 years.